Find the domain of $ y = \dfrac{{x - 1}}{{\sqrt {{x^2} - 1} }} $

Hint: The domain is the set of all possible and legitimate values of which can be put in a function without making it indeterminate. The term present in the square root is always greater than 0.

Complete step-by-step answer:
The given function is
 $ y = \dfrac{{x - 1}}{{\sqrt {{x^2} - 1} }} $
The critical points are considered by observing the denominator.
 $ \sqrt {{x^2} - 1} $
$\Rightarrow$ Always, $ {x^2} - 1 > 0 $
After observing the denominator it is clear that $ x = \pm 1 $ is not in the domain of the function as at
$ x = \pm 1 $ , the denominator becomes 0, which makes the function tending to infinity which is not required.
Also, by observation it is clear that we cannot put any value between $ \left[ { - 1,1} \right] $ as the square root term will be negative and square root function does not exist for negative values.
Thus, the domain of the given function is $ \left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right) $

Note: The important concept is that of a domain and properties of a square root function should be clear in mind before solving such problems.
For a square root function to be valid, the value of the function present inside the square root should always be positive i.e., greater than 0.
Domain is the set of legitimate values of function, which does not make the function as indeterminate.
For instance, for the function $ y = \dfrac{1}{{1 - x}} $ , the domain set includes all the real values of $ x $ except $ + 1 $ . Because at $ x = 1 $ the denominator tends to zero and the function goes to infinity. So, domain of the function is $ R - \left\{ 1 \right\} $